Let A,B,C,D be complex matrices $n \times n$ such that $AB^T,CD^T$ are symmetric and $AD^T-BC^T=I$. Prove that $A^TD-C^TB=I$. Can anyone give me any idea? Thank you.
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Is it intended to use only symetric matrices, not Hermitian? – TZakrevskiy Aug 09 '13 at 17:27
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yes only symetric matices – user62138 Aug 09 '13 at 17:31
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I'm very curious as to the origins of this problem. Can you provide some insights? – Robert Lewis Aug 09 '13 at 23:07
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Are you sure that there is no mistake in how you wrote the various relations? Because I believe I can show that what you are asked to prove cannot hold - and usually the assertion holds. – Alecos Papadopoulos Aug 10 '13 at 00:52
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I'm sure that there is no mistake.It's a exercise from a competition. – user62138 Aug 10 '13 at 06:57
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State what competition it is. If it is an ongoing competition, you are cheating by seeking help here. – user1551 Aug 10 '13 at 09:07
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No,it's of year 2003-2004 – user62138 Aug 10 '13 at 09:26
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Still, please state what competition it is. – user1551 Aug 10 '13 at 09:40
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admition test for a master degree – user62138 Aug 10 '13 at 09:47
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Also Putnam 1986-B6. – Catalin Zara Jan 05 '17 at 15:29
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Hint: The given condition says that $$ \pmatrix{A&-B\\ -C&D}\pmatrix{D^T&B^T\\ C^T&A^T} = \pmatrix{I&0\\ 0&I}. $$ Now, note that $XY=I$ implies that $YX=I$ and in turn $X^TY^T=I$.
user1551
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1Readers should note that the trick user1551 used of bundling operators on $H$ into operators on $H \oplus H$ can be quite handy in general, cf. the slick proof of Putnam's theorem from Fulgede's theorem (http://en.wikipedia.org/wiki/Fuglede's_theorem, Second Proof of Putnam's Theorem). – Branimir Ćaćić Aug 10 '13 at 09:53
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2The nicest example of such bundling that I have ever seen is given by N.S.'s answer in another thread. The classical proof that $AB$ and $BA$ have identical characteristic polynomials also uses such bundling. – user1551 Aug 10 '13 at 10:40